anonymous
  • anonymous
Why do some law of sines problems have none or two solutions?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Because a number canbe positive or negative and still result in the same outcome in a law of sines problem.
Kaederfds
  • Kaederfds
and it can be has more than 2 solutions too
anonymous
  • anonymous
can you explain better please...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Kaederfds
  • Kaederfds
because sin has periodic
jim_thompson5910
  • jim_thompson5910
This page may help you out http://www.regentsprep.org/Regents/math/algtrig/ATT12/lawofsinesAmbiguous.htm
anonymous
  • anonymous
because it depends on the quadrant of which the equation puts it. for example, sin is positive in the first and second quadrant (ASTC).. if your number is positive then it would be in quadrants I & II. if your number is negative then it would be in quadrants III & IV
anonymous
  • anonymous
jim do you agree with soccergurls answer?
jim_thompson5910
  • jim_thompson5910
yes, ATSC is an acronym that refers to you going from quadrant I to quadrant IV A: all are positive in quadrant 1 (sine, cosine, tangent) S: only sine is positive in quadrant 2 T: only tangent is positive in quadrant 3 C: only cosine is positive in quadrant 4 also, sine is positive for quadrants 1 & 2
jim_thompson5910
  • jim_thompson5910
and as the page I posted points out, sometimes you'll get something like sin(C) = 1.1428 but sin(x) is restricted from -1 to +1, which means sin(C) = 1.1428 has no solution and therefore, a triangle is not possible
anonymous
  • anonymous
ah, i think i get it now, thank you :)
jim_thompson5910
  • jim_thompson5910
yw, the page also goes over other examples too (of the 2 other cases)
anonymous
  • anonymous
okay, thank you :)
anonymous
  • anonymous
umm can i ask a follow up question...?
anonymous
  • anonymous
why does law of cosines not have any ambiguous cases?
jim_thompson5910
  • jim_thompson5910
To have an ambiguous case, it means that you aren't sure of what the angles are (if all the sides are known). Let's say we know the lengths of all 3 sides of two triangles. Furthermore, let's say that the corresponding lengths are congruent. If that's true, then we can use the SSS property to show that the two triangles are congruent. Now the question is: is it possible to have 2 congruent triangles but have one set of corresponding angles that are incongruent (of different measures)? The answer is no. The reason is the CPCTC property tells us that if we have congruent triangles, then the corresponding parts will be congruent as well. So the corresponding angles will be congruent, which means knowing all 3 sides leads you to find all 3 angles without any ambiguity or confusion. You can find any angle of a triangle using the law of cosines a^2 = b^2 + c^2 - 2bc*cos(A) where a,b,c are the sides and A,B,C are the angles (angle A is opposite side a, etc etc). So if you know a,b,c then A will be one fixed value and there won't be any ambiguity. This is true for B,C as well.
jim_thompson5910
  • jim_thompson5910
Unfortunately, if you're dealing with SSA, then you may run into ambiguity. This is because knowing two sides and the angle that's not between them doesn't uniquely determine a triangle (think back to geometry). So what this means is that if you were to plug in the sides and angle into a^2 = b^2 + c^2 - 2bc*cos(A), you would get a quadratic which has the potential to a) have 2 different solutions b) have exactly one solution c) have no solutions at all so the ambiguity still creeps up. But again, this is only possible if working with SSA

Looking for something else?

Not the answer you are looking for? Search for more explanations.